1.) If 2.86 moles of nitrogen reacts with 4.27 moles of hydrogen, what is the ma
ID: 1007375 • Letter: 1
Question
1.) If 2.86 moles of nitrogen reacts with 4.27 moles of hydrogen, what is the maximum volume of ammonia (in L) that can be produced at a pressure of 1.14 atm and 301 K.
N2(g) + 3 H2(g) 2 NH3(g)
2.) 13.4 g of butane (58.12 g/mol) undergoes combustion according to the following equation. What pressure of carbon dioxide in atm is produced at 306 K in a 1.52 L flask.
2 C4H10(g) + 13 O2 (g) 8 CO2 (g) + 10 H2O (g)
3.)
Order the gases from fastest (1.) to slowest (5.) molecular velocity at 25°C and 1 atm.
Kr H2 Ne N2 F2
Explanation / Answer
1)
N2(g) + 3 H2(g) ---> 2 NH3(g)
1 MOL n2 = 3 mol H2
No of mol of N2 = 2.86 mol
No of mol of H2 = 4.27 MOL
LIMITING REAGENT = H2
No of mol of NH3 produced = 4.27*2/3 = 2.847 mol
volume of NH3 = nRT/P = 2.847*0.0821*301/1.14 =61.71 L
2) 2 C4H10 (g) + 13 O2 (g) 8 CO2 (g) + 10 H2O (g)
No of mol of butane = 13.4/58 = 0.231 mol
No of mol ofCO2 produced = 0.231*8/2 = 0.924 mol
PRESSURE of CO2 = 0.924*0.0821*306/1.52 = 15.27 atm
3) Kr < F2 < N2 < H2
molarmass increases, velocity decreases
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